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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Proof of the Method of Undetermined Coefficients</dfn>(i) Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html">
\begin{equation}
y^{\prime \prime}+b y^{\prime}+c y=P_n(x)=a_0 x^n+a_1 x^{n-1}+\cdots+a_{n-1} x+a_n.\tag{3.6.6}
\end{equation}
</div>
<p class="continuation">Assume that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html">
\begin{equation}
Y=A_0 x^n+A_1 x^{n-1}+\cdots+A_{n-1} x+A_n.\tag{3.6.7}
\end{equation}
</div>
<p class="continuation">Substituting it into the ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html">
\begin{equation}
\begin{aligned}
&amp;n (n-1) A_0 x^{n-2}+(n-1) (n-2) A_1 x^{n-3}+\cdots+2 A_{n-2}\\
&amp;\quad  \quad +b(n A_0 x^{n-1}+(n-1)A_1 x^{n-2}+\cdots+A_{n-1})\\
&amp;\quad  \quad  \quad+ c(A_0 x^n+A_1 x^{n-1}\cdots+A_{n-1} x+A_n)\\
&amp;\quad  \quad  \quad  \quad=a_0 x^n+a_1 x^{n-1}+\cdots+a_{n-1} x+a_n.
\end{aligned}\tag{3.6.8}
\end{equation}
</div>
<p class="continuation">After arrangement of the two sides, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html">
\begin{equation*}
\begin{cases}
c A_0=a_0\\
c A_1+ n b A_0=a_1\\
\cdots\\
c A_{n-1}+2 b A_{n-2} +6 A_{n-3}=a_{n-1}\\
c A_n+ b A_{n-1}+2 A_{n-2}=a_n
\end{cases}
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">If</dfn> <span class="process-math">\(c \neq 0\text{,}\)</span> <span class="process-math">\(A_0, A_1, \cdots, A_n\)</span> can be determined and this method works.<dfn class="terminology">If</dfn> <span class="process-math">\(c=0\)</span> but <span class="process-math">\(b \neq 0\text{,}\)</span> then the left hand side of (<a href="" class="xref" data-knowl="./knowl/eq3_16.html" title="Equation 3.6.8">(3.6.8)</a>) is a polynomial of degree <span class="process-math">\(n-1\text{,}\)</span> while the right hand side is always a polynomial of degree <span class="process-math">\(n\text{.}\)</span> Thus (<a href="" class="xref" data-knowl="./knowl/eq3_16.html" title="Equation 3.6.8">(3.6.8)</a>) can never be satisfied. To make the left hand side of (<a href="" class="xref" data-knowl="./knowl/eq3_16.html" title="Equation 3.6.8">(3.6.8)</a>) to have degree <span class="process-math">\(n\text{,}\)</span> we have to assume that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html">
\begin{equation}
Y=x (A_0 x ^n+A_1 x^{n-1}+\cdots+A_{n-1} x+A_n).\tag{3.6.9}
\end{equation}
</div>
<p class="continuation"><dfn class="terminology">If</dfn> both <span class="process-math">\(b=0\)</span> and <span class="process-math">\(c=0\text{,}\)</span> then the left hand side of (<a href="" class="xref" data-knowl="./knowl/eq3_16.html" title="Equation 3.6.8">(3.6.8)</a>) has degree <span class="process-math">\(n-2\text{.}\)</span> Thus, we have to assume</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html ./knowl/eq3_16.html">
\begin{equation}
Y=x^2 (A_0 x ^n+A_1 x^{n-1}+\cdots+A_{n-1} x+A_n).\tag{3.6.10}
\end{equation}
</div>
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